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Calculus IChapter 4: Application of DerivativesSection 4-14

Business Applications — Question 8

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Question 8

The weekly revenue (in thousands of dollars) for producing xx units is: R(x)=120x−x2R(x) = 120x - x^2 and the weekly cost is: C(x)=40x+200C(x) = 40x + 200

  • (a) Determine the profit function P(x)P(x).

  • (b) Find the number of units xx that maximizes profit.

  • (c) Compute the maximum profit.

Original worksheet page 1: question and worked solution for 4-14-008
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Question 8 - Solution

(a) Profit Function: P(x)=R(x)−C(x)=(120x−x2)−(40x+200)=−x2+80x−200P(x) = R(x) - C(x) = (120x - x^2) - (40x + 200) = -x^2 + 80x - 200

(b) Maximize Profit: P′(x)=−2x+80P'(x) = -2x + 80

Set P′(x)=0P'(x) = 0: −2x+80=0⇒x=40-2x + 80 = 0 \quad \Rightarrow \quad x = 40

Second derivative: P″(x)=−2<0P''(x) = -2 < 0 Hence, profit is maximized when x=40x = 40.

(c) Maximum Profit: P(40)=−(40)2+80(40)−200=−1600+3200−200=1400P(40) = -(40)^2 + 80(40) - 200 = -1600 + 3200 - 200 = 1400

Maximum profit occurs at x=40 units, Pmax=1400 thousand dollars\boxed{\text{Maximum profit occurs at } x = 40 \text{ units, } P_{\max} = 1400 \text{ thousand dollars}}

Original worksheet page 2: question and worked solution for 4-14-008

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